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$\displaystyle\int_1^e (\ln x)^2\, dx = ?$

A$e - 2$
B$e + 2$
C$e^2 - 1$
D$2e - 1$
Answer & Solution
Correct answer: A. $e - 2$
**IBP:** $u = (\ln x)^2$, $dv = dx$. $= [x(\ln x)^2]_1^e - \int_1^e 2\ln x\, dx = e \cdot 1 - 0 - 2[x\ln x - x]_1^e = e - 2[(e - e) - (0 - 1)] = e - 2(1) = e - 2$.
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